A 4.0 cm long straight wire carrying a current of 8 A is placed perpendicular to a uniform magnetic field of strength 0.15 T . The magnetic force on the wire is ________mN .

<p>To calculate the magnetic force on a wire, we use the formula:</p> <p>$ \mathrm{F} = \mathrm{I} \ell \mathrm{B} $</p> <p>Where:</p> <p><p>$ \mathrm{I} $ is the current in the wire (in amperes),</p></p> <p><p>$ \ell $ is the length of the wire (in meters),</p></p> <p><p>$\mathrm{B}$ is the magnetic field strength (in teslas).</p></p> <p>Given that:</p> <p><p>The current $\mathrm{I}$ is 8 A,</p></p> <p><p>The length of the wire $\ell$ is 4.0 cm, which is 0.04 m (since 1 cm = 0.01 m),</p></p> <p><p>The magnetic field strength $\mathrm{B}$ is 0.15 T,</p></p> <p>Substituting these values into the formula gives:</p> <p>$ \mathrm{F} = 8 \times 0.04 \times 0.15 $</p> <p>Calculating this:</p> <p>$ \mathrm{F} = 0.048 \, \mathrm{N} $</p> <p>To convert this to millinewtons (mN), recall that $1 \, \mathrm{N} = 1000 \, \mathrm{mN}$. Therefore:</p> <p>$ \mathrm{F} = 48 \, \mathrm{mN} $</p> <p>Thus, the magnetic force on the wire is 48 mN.</p>